**2. Fin and tube heat exchanger model**

### **2.1 Numerical model**

Our numerical model couples with CFD for air-side calculations and with the onedimensional refrigerant circuit model for refrigerant-side calculations. The air-side model consist of two CFD models. The first model calculates airflow velocity and pressure by solving an unsteady Reynolds-averaged Navier–Stokes (RANS) equation model. The second model calculates airflow temperature and amount of heat exchange of each tube by using first model calculation result. This model solves a steady heat conductivity equation where the airflow velocity field calculation result is fixed. The refrigerant flow side model can calculate refrigerant flow pressure, enthalpy, and flow rates on each tube by using circuit model of a heat exchanger path. This model inputs heat of tubes from the air-side model result. **Table 1** denotes the input and output relationship of these models.

#### *2.1.1 Air-side model*

A commercial CFD software, SCRYU/Tetra (CRADLE 2018) [12], was used for airside models. Two air-side CFD models, airflow model, and heat conductivity model use same CFD mesh. **Figure 4** shows the CFD mesh of the overall heat exchanger and the cross-flow fan, and **Figure 4b** shows the fin surface mesh. This mesh consists of a

*Application of CFD to Prediction of Heat Exchanger Temperature and Indoor Airflow Control… DOI: http://dx.doi.org/10.5772/intechopen.110076*


#### **Table 1.**

*Fin-and-tube heat exchanger model.*

#### **Figure 4.**

*Mesh of indoor unit: (a) internal of indoor unit and (b) fins and tubes. Tube walls are boundary condition regions of CFD (heat conductivity) model, which can be set to temperature from the result of refrigerant flow-side model.*

stationary region mesh and a rotating region mesh. For rotating region, we created 0.25 mm to 0.50 mm tetrahedral meshes. Four prism meshes of 0.1 mm are inserted along the fan surface. Arbitrary Lagrangian Euler (ALE) method is adopted for fan rotating calculation. A tetrahedral unstructured mesh size was 0.25 mm to 50 mm and further prism layers were inserted along the walls. For fin walls, tetrahedral mesh size is 0.25 mm and 0.1 mm; two prism layers were inserted along the walls. Actual typical indoor unit heat exchanger width is 666 mmm and has about 555 fins. In order to limit the number of meshes and perform calculations efficiently within our computational resource, we designed our numerical model to have width of 6.0 mm and five fins. Previous research papers [2, 5] used two dimensional CFD model for simulating crossflow fan airflow pattern. We used thin CFD model (6.0 mm width model) that models a the cross-sectional view at a representative position. For the thin width, this model airflow pattern is two dimensional. However, the actual indoor unit has filters and unit ribs that block the airflow at some positions. The cross-sectional view of the indoor unit has different shapes in different positions and, therefore, airflow patterns are different at these positions. When calculating the amount of heat exchange for each tube, the results are multiplied by 100.0, which is an adjusted constant to consider the influence of some different shapes of cross-sectional view of the actual unit.

#### *2.1.2 Refrigerant flow-side model*

We constructed a circuit model, which is made of node elements and tube elements to solve refrigerant flow state (**Figure 5a**). The node element preserves

**Figure 5.**

*Refrigerant flow circuit model: (a) arrangement of node and tube elements in the heat exchanger and (b) a relationship of the node and tube elements.*

refrigerant flow temperature T (°C), enthalpy H (kJ/kg), and pressure P (Pa). The tube element is connected to two node elements. It preserves the tube length l (m), tube inner diameter *ϕ* (m), the refrigerant flow rate G (kg/s), the amount of heat transfer between refrigerant flow and the air *qair* (W), and tube wall temperature *Ttube*. The distributor can be modeled with more than three node elements with tube elements. **Figure 5b** shows the relationship of the node and tube elements, indicated by the two nodes: Node (i) and Node (j), the tube elements connecting with Node (i) and Node (j) represented by Tube (ij).

Let *Pi* be the element pressure of Node (i), *Ti* be the temperature, and *Hi* be the enthalpy. The difference of Node (i) pressure *Pi* and Node (j) pressure *Pj* can be written by using the function <sup>Δ</sup>*P Gij*, *<sup>ϕ</sup>ij*, *lij* � �, which calculates pressure loss from the Tube (ij) refrigerant flow rate *Gij* using Blasius equation and Lockhart-Martinelli correlation [3]. The relationship of *Pi*, *Pj*, and <sup>Δ</sup>*P Gij*, *<sup>ϕ</sup>ij*, *lij* � � is as in Eq. (1).

$$P\_i - P\_j = \Delta P(G, \phi, l) \tag{1}$$

The relationship of the connecting node elements Node (i) and Node (j), enthalpy *Hi*, *Hj*, and *Gij* is shown in Eq. (2).

$$H\_i - H\_j = q\_{a \dot{r}, \dot{\imath}j} / \mathcal{G}\_{i,j} \tag{2}$$

Suppose that the refrigeration circuit reaches the steady state, at any node element Node (j), the total refrigerant flow rate is equal to 0 (kg/s).

$$\sum\_{j=1} G\_{i,j} = \mathbf{0} \tag{3}$$

Eq. (1) is the momentum conservation equation, Eq. (2) is the energy conservation equation, and Eq. (3) is the mass conservation equation.

Assuming that the number of node elements is *n* and the number of tube elements is *m*, the number of unknown variables is *n* each for *Pi* and *Hi*, and *m* for *Gij*, for a total unknown variable is 2*n* þ *m*. The number of equations is *n* for each of Eqs. (2) and (3), and m for Eq. (3), for a total number of equations is 2*n* þ *m*. Since the number of unknown variables and the number of equations are equal, the value of the unknown

*Application of CFD to Prediction of Heat Exchanger Temperature and Indoor Airflow Control… DOI: http://dx.doi.org/10.5772/intechopen.110076*

variable can be uniquely determined from equations Eq. (1), (2), and (3). **Figure 6** is a heat exchanger calculation flowchart. We can obtain refrigerant flow *Pi* (Pa) and enthalpy *Hi* kJ/kg for all node elements.

We can calculate temperature of node elements *Ti* from *Pi* and *Hi* by using refrigerant property database REFPROP [13]. We introduce the refrigerant flow temperature of *Tubeij* element as *Tref* ¼ *Ti* þ *Tj =*2 (°C). We can then calculate wall temperature *Ttube*,*ij* from the heat transfer rate of tube *αij* (W/m<sup>2</sup> K), tube surface *Sij* (m<sup>2</sup> ), tube wall thickness t (m), and tube wall heat conductivity (W/m2 K) as follows:

$$T\_{tube, \vec{ij}} = T\_{ref} + \left(\frac{\mathbf{1}}{\alpha\_{\vec{ij}} \mathfrak{s}\_{\vec{ij}}} + \frac{t}{\lambda}\right)^{-1} q\_{air, \vec{ij}} \tag{4}$$

#### *2.1.3 Coupling calculation*

**Figure 7** is a flowchart of coupling calculation. Initially, the air-side CFD (airflow) is performed once. Then each tube wall temperature *Ttube*,*ij* is assumed as the initial condition and CFD (heat conductivity) process is performed. This CFD (heat conductivity) process outputs each tube *qair*,*ij*. Third, the refrigeration circuit simulation process calculates the refrigerant flow state using *qair*,*ij* and outputs *Ttube*,*ij*, which are different from the CFD initial condition *Ttube*,*ij*. The second and third process is an iterative loop processes.

In the previous research [5], it was necessary to fix the temperature of the tubes *Ttube*,*ij*i to perform the CFD (heat conductivity). In addition, it was not possible to calculate refrigerant enthalpy. In this coupled calculation, the heat transfer from the air to the refrigerant flow is the same value (*qair*,*ij*) for both the air-side and refrigerant flow calculations. This makes it possible to calculate the refrigerant enthalpy at the same time the temperature of tubes is calculated iteratively.

#### **Figure 7.** *Flowchart of coupled simulation.*


**Table 2.** *Air-side CFD result.*
